12. Two bulbs of 100 W each and two coolers of 250 W each, work on an average 6 hours a day. If the energy
costs Rs. 1-75 per kWh, calculate the monthly bill and the minimum fuse rating when power is supplied
at 250 V.
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Answer:
the monthly bill and the maximum fuse rating when power is supplied at 250v is Rs 6.475
Explanation:
solution:
we know that,
power p =energy/time
energy=power/time
2 bulbs of 100 w each,
100 ₓ 2 = 200 w per day
two coolers of 250 w each,
250 ₓ 2 = 500 w per day ,
The bulbs are working nearly 6 hours per day
so, the energy consumed in 1 day
for bulb = 200 ₓ 6 = 1200 key word optimized for developer = 1.2 KW/h
For cooler =500 ₓ 5 = 2500 KW/h =2.5 KW/h
Total energy consumed (add bulb and coolers watt)=3.7 KW/h
1 unit = 1 KW/h
1 unit costs Rs 1.75
on calculating 3.7 units,we get
3.7 ₓ 1.75 =Rs 6.475 (monthly bill
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